Description#
Given a positive integer num
represented as a string, return the integer num
without trailing zeros as a string.
Example 1:
Input: num = "51230100"
Output: "512301"
Explanation: Integer "51230100" has 2 trailing zeros, we remove them and return integer "512301".
Example 2:
Input: num = "123"
Output: "123"
Explanation: Integer "123" has no trailing zeros, we return integer "123".
Constraints:
1 <= num.length <= 1000
num
consists of only digits.num
doesn't have any leading zeros.
Solutions#
Solution 1#
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| class Solution:
def removeTrailingZeros(self, num: str) -> str:
return num.rstrip("0")
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| class Solution {
public String removeTrailingZeros(String num) {
int i = num.length() - 1;
while (num.charAt(i) == '0') {
--i;
}
return num.substring(0, i + 1);
}
}
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| class Solution {
public:
string removeTrailingZeros(string num) {
while (num.back() == '0') {
num.pop_back();
}
return num;
}
};
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| func removeTrailingZeros(num string) string {
i := len(num) - 1
for num[i] == '0' {
i--
}
return num[:i+1]
}
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| function removeTrailingZeros(num: string): string {
let i = num.length - 1;
while (num[i] === '0') {
--i;
}
return num.substring(0, i + 1);
}
|
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| impl Solution {
pub fn remove_trailing_zeros(num: String) -> String {
let mut i = num.len() - 1;
while num.chars().nth(i) == Some('0') {
i -= 1;
}
num[..i + 1].to_string()
}
}
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Solution 2#
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| impl Solution {
pub fn remove_trailing_zeros(num: String) -> String {
num.chars()
.rev()
.skip_while(|&c| c == '0')
.collect::<String>()
.chars()
.rev()
.collect::<String>()
}
}
|